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PHYSICS DEMONSTARTION PROJECT Made by : Jaskaran Singh Cheema Class : XII B Roll no. : 09 Session : 2012-2013
CERTIFICATE This is to certify that Jaskaran Singh Cheema, Roll number _________ of class XII-D has successfully completed the PHYSICS DEMONSTRATION EXPERIMENTS under my supervision according to the guidelines laid down by CBSE.
TEACHER INCHARGE
ACKNOWLEGDEMENTS I acknowledge the invaluable contribution of my physics teacher, Mrs. Jyoti Nanda(Physics Dept.) in helping me to complete these demonstrations. These experiments wouldn’t have been possible to complete without her moral support and proper guidance. Also I’m deeply grateful to Mr. Joseph for his constant help. Jaskaran Singh Cheema
CONTENTS 1) DEMONSTRATION •
Aim
•
Apparatusrequired
•
Diffraction
•
Observationtable
•
Conclusion
•
Biblography
DEMONSTRATION AIM: To observe the effect on the width of central maxima by changing the distance between screen and the slit in diffraction. APPARATUS REQUIRED: laser, single slit, opticalbench, screen, scale. THEORY:
DIFFRACTION Diffraction is the phenomenon of bending of light at the sharp edge of the obstacle.
Condition for diffraction: The width of the obstacle should be comparable with the wavelength of light falling on the obstacle.
Single slit experiment: Path difference=BC=asin0 In diffraction light waves come front two
different parts of same wave front superimpose and alternate dark and bright bands are obtained on this screen.The intensity of light at the centre is maximum and goes on decreasing from two sides of central maxima so other maxima and minima are called secondary maxima and secondary minima.
CONDITION FOR SECONDARY MINIMA:1. Path difference = n asin0 = n where n = 1,2,3,......... It is because the waves coming from two different parts of same wavefront have the path difference / 2 or multiple of / 2 2. Phase difference= 0 = 2Nii where n=1,2,3...... 3. Position of n dark band or n secondary minima = x = n D/a
CONDITION FOR SECONDARY MAXIMA:1. Path difference= (2n+1) /2 asin0 = (2n+1) /2 where n =1,2,3...... It is because when the path difference between the two waves coming from two different parts is the multiple of and / 2respectively and maxima is obtained. 2. Phase difference = 0 = (2n+1)ii 3. Position of n bright band or n secondary maxima x =(2n+1) D/2a
EXPRESSION FOR WIDTH OF CENTRAL MAXIMA IN SINGLE SLIT EXPERIMENT:
It is the distance between the first two secondary minima on both sides of the central point.
For secondary minima •asin0
= n •For n =1 •asin0 = •So, sin0 = /a .........................(1) •And,
sin0 = x/a •From (1) and (2) •Sin0= /a=x/a •x = D/a ......................(2) Hence, width of the central minima is x = D/a
WIDTH OF CENTRAL MAXIMA •B
= 2x •B=2 D/a
OBSERVATION TABLE:-
CONCLUSION:Width of central maxima increases with increase in distance between the slit and the screen.
BIBLIOGRAPHY Wikipedia – The Free Encyclopedia Physics NCERT Class XII Textbook of Physics –
Pradeep’s Encarta Encyclopedia Britannica Encyclopedia
